Statistics with Grouped Data
Senior 2 · Mathematics
Statistics with Grouped Data
Lesson 3 of 5
20 min
Reports rarely give you every raw value. A district education office might publish only a grouped table of students' travel times or test marks. You can still find a very good estimate of the mean, the "fair share" average, from the table alone.
Inside a class we no longer know the exact values. So we assume every value in a class sits at the class midpoint. Then each class contributes (midpoint × frequency) to the total, and
mean ≈ Σfx ÷ Σf
where x is the class midpoint, f the frequency and Σ ("sigma") means "add them all up".
| Time (min) | Midpoint x | Frequency f | fx |
|---|---|---|---|
| 5–14 | 9.5 | 5 | 47.5 |
| 15–24 | 19.5 | 12 | 234 |
| 25–34 | 29.5 | 13 | 383.5 |
| 35–44 | 39.5 | 7 | 276.5 |
| 45–54 | 49.5 | 3 | 148.5 |
| Total | Σf = 40 | Σfx = 1090 |
Mean ≈ 1090 ÷ 40 = 27.25 minutes.
The exact mean of the 40 raw values is 27.025 minutes. The grouped estimate is close but not identical, because some values are below their midpoint and some above. That is why we say "estimate".
Daily milk yield (litres) of 20 cows at a cooperative in Nyagatare:
| Litres | 0 ≤ x < 4 | 4 ≤ x < 8 | 8 ≤ x < 12 | 12 ≤ x < 16 |
|---|---|---|---|---|
| Cows | 3 | 8 | 6 | 3 |
Midpoints: 2, 6, 10, 14. Σfx = 3 × 2 + 8 × 6 + 6 × 10 + 3 × 14 = 6 + 48 + 60 + 42 = 156. Mean ≈ 156 ÷ 20 = 7.8 litres.
Tip
Divide by Σf, not by the number of classes. In Worked example 1, 1090 ÷ 5 = 218 minutes is absurd: the mean must lie between the smallest and largest class. Always sanity-check your mean against the table.
Marks out of 50 of 30 students: 0–9: 2; 10–19: 5; 20–29: 9; 30–39: 10; 40–49: 4. Estimate the mean mark.
Solution: midpoints 4.5, 14.5, 24.5, 34.5, 44.5. Σfx = 9 + 72.5 + 220.5 + 345 + 178 = 825. Mean ≈ 825 ÷ 30 = 27.5 marks.
Why is the answer to question 1 only an estimate?
Solution: we replaced every mark by its class midpoint; the real marks are spread through each class.
A student divides Σfx = 825 by 5 and gets 165. What went wrong?
Solution: 5 is the number of classes; the mean needs the number of students, Σf = 30. A mean of 165 marks out of 50 is impossible.
Using lower class limits instead of midpoints in question 1, would the estimate be too high or too low?
Solution: too low, because every value would be pushed to the bottom of its class.
Remember
Sources and credits
The Rwandan examples, practice items and exercises are written by Ganzaa.
